CSP-J 2023 T3 一元二次方程 解题报告
前言
今年\(CSP\)的原题, 回家\(1h\)内写\(AC\), 但是考场上没有写出来 , 原因是脑子太不好了, 竟然调了两个小时没有调出来. 一等奖悬那......
正题
看完题目,第一眼就是大模拟, 并且\(CCF\)绝对不会让你好受,所以出了一个如此***钻的题目, 并且要考虑到非常多的情况, 代码非常长......
最重要的一点: \(\Delta\)
\(\Delta\)是此题中最重要的分情况讨论的地方. 根据初三\(22\)章的所学知识 ,可知分三种情况:
1. \(\Delta < 0\)
不用说了
直接输出NO
2. \(\Delta = 0\)
同样的, 只有一种情况, 答案为\(- \dfrac{b}{2a}\),但是, 需要严谨的判断.
if(delta == 0) {
if(b == 0) {
cout << 0;
}
else if(a * b > 0) {
a = abs(a);
b = abs(b);
cout << "-";
if(b % (2 * a) == 0) {
cout << b / 2 / a;
}
else {
cout << b / __gcd(2 * a, b) << "/" << (2 * a) / __gcd(2 * a, b);
}
}
else {
a = abs(a);
b = abs(b);
if(b % (2 * a) == 0) {
cout << b / 2 / a;
}
else {
cout << b / __gcd(2 * a, b) << "/" << (2 * a) / __gcd(2 * a, b);
}
}
}
3. \(\Delta > 0\)
地狱.
我是分两种情况的, 一种是\(a > 0\), 一种是\(a < 0\). 这样可以分辨出是\(+ \sqrt{\Delta }\) 还是\(-\sqrt{\Delta }\)
如若\(a < 0\), 则可知答案为:
\[\dfrac{b + \sqrt{\Delta}}{-2a} \]如若\(a > 0\), 则可知答案为:
\[\dfrac{\sqrt{\Delta} - b}{2a} \]- 在这里有一个技巧, 就是不论怎样, 输出时, \(\sqrt{\Delta}\)永远是正的(符号为
+)
可以分两种情况:
1.第一种: 不需要写sqrt, 也就是\(\Delta\)为完全平方数时,
比较好处理, 首先需要判断\(b + \sqrt{\Delta}\)是否为\(0\). 如果是, 则直接输出\(0\); 否则输出最简分数.
其中, 一定要记住如果\((b + \sqrt{\Delta}) \% (2 * a) = 0\), 就直接输出一个整数.还要注意判断正负号.
2.第二种: 需要写sqrt, 很难.
首先, 先输出前面的内容, 也就是\(-\dfrac{b}{2a}\), 判断同上.
然后, 输出+, 代表符号.
接着, 找出三个变量, 也就是: \(\dfrac{x}{y} \sqrt{\dfrac{\Delta}{x^2}}\)中的\(x, y和\dfrac{\Delta}{x^2}\).其中,\(\sqrt{\dfrac{\Delta}{x^2}}\)为最简平方根数.
接下来是\(4\)种情况:
\(x = y\), 只有\(\sqrt{\dfrac{\Delta}{x^2}}\);
\(x \% y = 0\), 只有\(\dfrac{x}{y}\sqrt{\dfrac{\Delta}{x^2}}\)
\(y \% x = 0\), 只有\(\dfrac{\sqrt{\dfrac{\Delta}{x^2}}}{y}\)
其他情况, 输出\(\dfrac{x \times \sqrt{\dfrac{\Delta}{x^2}}}{y}\)
完结撒花!!
\(Code:\)
- 心脏病患者请勿观看
#include <bits/stdc++.h>
using namespace std;
int T, M;
int a, b, c;
int pd(int x) {
for(int i = sqrt(x) + 1; i >= 1; --i) {
if(x % (i * i) == 0) {
return i;
}
}
}
int main() {
cin >> T >> M;
while(T--) {
cin >> a >> b >> c;
int delta;
delta = b * b - 4 * a * c;
if(delta < 0) {
cout << "NO";
}
else if(delta == 0) {
if(b == 0) {
cout << 0;
}
else if(a * b > 0) {
a = abs(a);
b = abs(b);
cout << "-";
if(b % (2 * a) == 0) {
cout << b / 2 / a;
}
else {
cout << b / __gcd(2 * a, b) << "/" << (2 * a) / __gcd(2 * a, b);
}
}
else {
a = abs(a);
b = abs(b);
if(b % (2 * a) == 0) {
cout << b / 2 / a;
}
else {
cout << b / __gcd(2 * a, b) << "/" << (2 * a) / __gcd(2 * a, b);
}
}
}
else {
if(a < 0) {
int mother = - 2 * a;
int x = pd(delta);
int y = delta / x / x;
if(b == 0) {
mother = abs(mother);
if(y == 1) {
if(x == mother) {
cout << "1";
}
else if(x % mother == 0) {
cout << x / mother;
}
else {
cout << x / __gcd(x, mother) << "/" << mother / __gcd(x, mother);
}
}
else {
if(x == mother) {
cout << "sqrt(" << y << ")";
}
else if(mother % x == 0) {
cout << "sqrt(" << y << ")";
cout << "/" << mother / x;
}
else if(x % mother == 0) {
cout << x / mother << "*sqrt(" << y << ")";
}
else {
cout << x / __gcd(x, mother) << "*sqrt(" << y << ")" << "/" << mother / __gcd(x, mother);
}
}
}
else if(y == 1) { // 不需要sqrt
// 说明可以合并为同一个式子
int son = - b - x;
if(son == 0) {
cout << 0;
}
else if(son * mother < 0) { // 如果分子分母同号.
son = abs(son);
mother = abs(mother);
if(son % mother == 0) {
cout << son / mother;
}
else {
cout << son / __gcd(son, mother) << "/" << mother / __gcd(son, mother);
}
}
else { // 如果分子分母异号.
son = abs(son);
mother = abs(mother);
cout << "-";
if(son % mother == 0) {
cout << son / mother;
}
else {
cout << son / __gcd(son, mother) << "/" << mother / __gcd(son, mother);
}
}
}
else { // 需要sqrt.
// 1. 先输出前面的
if(b > 0) { // 不需要负号
b = abs(b);
mother = abs(mother);
if(b % mother == 0) {
cout << b / mother;
}
else {
cout << b / __gcd(b, mother) << "/" << mother / __gcd(b, mother);
}
}
else { // 需要负号
b = abs(b);
mother = abs(mother);
cout << "-";
if(b % mother == 0) {
cout << b / mother;
}
else {
cout << b / __gcd(b, mother) << "/" << mother / __gcd(b, mother);
}
}
// 2. 输出sqrt部分(不管怎样都是+)
cout << "+";
if(x == 1) { // 不需要输出前缀.
cout << "sqrt(" << y << ")";
cout << "/" << - 2 * a;
}
else {
if(x == mother) {
cout << "sqrt(" << y << ")";
}
else if(x % mother == 0) {
cout << x / mother << "*sqrt(" << y << ")";
}
else if(mother % x == 0) {
cout << "sqrt(" << y << ")";
cout << "/" << mother / x;
}
else {
cout << x / __gcd(x, mother);
cout << "*sqrt(" << y << ")";
cout << "/" << mother / __gcd(x, mother);
}
}
}
}
else {
int mother = 2 * a;
int x = pd(delta);
int y = delta / x / x;
if(b == 0) {
mother = abs(mother);
if(y == 1) {
if(x == mother) {
cout << "1";
}
else if(x % mother == 0) {
cout << x / mother;
}
else {
cout << x / __gcd(x, mother) << "/" << mother / __gcd(x, mother);
}
}
else {
if(x == mother) {
cout << "sqrt(" << y << ")";
}
else if(mother % x == 0) {
cout << "sqrt(" << y << ")";
cout << "/" << mother / x;
}
else if(x % mother == 0) {
cout << x / mother << "*sqrt(" << y << ")";
}
else {
cout << x / __gcd(x, mother) << "*sqrt(" << y << ")" << "/" << mother / __gcd(x, mother);
}
}
}
else if(y == 1) { // 不需要sqrt
// 说明可以合并为同一个式子
int son = - b + x;
if(son == 0) {
cout << 0;
}
else if(son * mother > 0) { // 如果分子分母同号.
son = abs(son);
mother = abs(mother);
if(son % mother == 0) {
cout << son / mother;
}
else {
cout << son / __gcd(son, mother) << "/" << mother / __gcd(son, mother);
}
}
else { // 如果分子分母异号.
son = abs(son);
mother = abs(mother);
cout << "-";
if(son % mother == 0) {
cout << son / mother;
}
else {
cout << son / __gcd(son, mother) << "/" << mother / __gcd(son, mother);
}
}
}
else { // 需要sqrt.
// 1. 先输出前面的
if(b * mother < 0) { // 不需要负号
b = abs(b);
mother = abs(mother);
if(b % mother == 0) {
cout << b / mother;
}
else {
cout << b / __gcd(b, mother) << "/" << mother / __gcd(b, mother);
}
}
else { // 需要负号
b = abs(b);
mother = abs(mother);
cout << "-";
if(b % mother == 0) {
cout << b / mother;
}
else {
cout << b / __gcd(b, mother) << "/" << mother / __gcd(b, mother);
}
}
// 2. 输出sqrt部分(不管怎样都是+)
cout << "+";
if(x == 1) { // 不需要输出前缀.
cout << "sqrt(" << y << ")";
cout << "/" << 2 * a;
}
else {
mother = 2 * a;
if(x == mother) {
cout << "sqrt(" << y << ")";
}
else if(x % mother == 0) {
cout << x / mother << "*sqrt(" << y << ")";
}
else if(mother % x == 0) {
cout << "sqrt(" << y << ")";
cout << "/" << mother / x;
}
else {
cout << x / __gcd(x, mother);
cout << "*sqrt(" << y << ")";
cout << "/" << mother / __gcd(x, mother);
}
}
}
}
}
cout << endl;
}
return 0;
}